What are stamens?

Explanation

We are given that
Heat of combustion of `CH_(3) COOH = - 867 kj/mol`
(i) `CH_(3) COOH+2O_(2) to 2CO_(2) + 2H_(2) O, Delta H = - 867 kj mol^(-1)`
Heat of formation of `CO_(2) = - 393.5 kj mol^(-1)`
Heat of formation of `CO_(2) = - 393.5 kj mol^(-1)`
(ii)`C + O_(2) to CO_(2) (g), Delta H = 393.5 kj mol^(-1)`
Heat of formation of `H_(2)O = - 285.9 kj mol^(-1)`
(iii) `H_(2) + (1)/(2) O_(2) to H_(2) O (l), Delta H = - 285.9 kj mol^(-1)`
We are to calulate the heat of the formation `(Delta H)` of acetic acid. This can be done by solving above equations a follows : `2C + 2H_(2) + O_(2) to CH_(3)COOH`
Multiplying equation (ii) and (iii) by 2, we get
(iv) `2 C +2O_(2) to 2CO_(2) (g), Delta H = - 787 kj mol^(-1)`
(v) `2H_(2) + O_(2) to 2H_(2)O(l) , Delta H = - 571.8 kj mol^(-1)`
Now by adding equation (iv) and (v), we get
`2C + 2O_(2) to 2CO_(2) (g), Delta H = - 787.0 kj mol^(-1)`
(vi) `2H_(2) + O_(2) to 2H_(2)O (l) , Delta H = - 571.8 kj mol^(-1)`
`2C + 2H_(2) + 3O_(2) to 2CO_(2)+2H_(2)O,Delta H = - 1358.8 kj mol^(-1)`
Now subtract equation (i) from (iv), we get
`2C + 2H_(2) + 3O_(2) to 2CO_(2) + 2H_(2)O,Delta H = - 1358.8 kj mol^(-1)`
`CH_(3) COOH + 2O_(2) to 2CO_(2) + 2H_(2) O,Delta H = - 867 kj mol^(-1)`
`2C + 2H_(2) + 3O_(2) to CH_(3)COOH,Delta H = - 491.8 km mol^(-1)`